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Assuming the Riemann Hypothesis,

$$\pi(x)=R(x)-\sum_\rho R(x^\rho),$$

where $\pi(x)$ is the prime counting function and $R(x)$ is the Riemann prime counting function.

What does a plot of $g(x)=e^{\pi'(x)}$ look like?

Another option might be using:

$$\quad\Pi'(x)=\sum\limits_{n=2}^\infty\frac{\Lambda(n)}{\log(n)}\,\delta(x-n)\quad\text{(derivative of Riemann's prime-power counting function)}.$$

Define $f(x)=e^{\Pi'(x)}.$ What does the plot of $f(x)$ look like?

Edit: Assuming the Riemann Hypothesis $\pi(x)=Li(x)+O(\sqrt{x}\ln(x)).$

I can take the derivative of the first term $Li(x)$ but I'm not sure how to take the derivative of the second term.

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    $\begingroup$ No need to assume the RH. Replace $\pi(x)$ by $\psi(x)$ and its corresponding explicit formula $\psi(x) = (x-\sum_\rho x^\rho/\rho-C-\sum_{k=1}^\infty x^{-2k}/(-2k))1_{x > 1}$ which is much easier, $\psi'(x) = \sum_n \Lambda(n) \delta(x-n)$ is a distribution, $e^{\delta(x)}$ and $e^{\psi'(x)} $ don't make any sense. If your question is if $(1-\sum_\rho x^{\rho-1}-\sum_{k=1}^\infty x^{-2k-1})1_{x > 1}$ (dont forget the $1_{x > 1}$) converges to the distribution $\psi'(x)$ then yes this is the Weil explicit formula. It doesn't converge in the function sense. $\endgroup$ – reuns May 5 at 16:50
  • $\begingroup$ @Ultradark I believe the answer I posted to a related question at math.stackexchange.com/q/3186899 perhaps provides some insight. $\endgroup$ – Steven Clark May 5 at 19:58

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